MotionInCentralForceField 2009-03-31 13:58:55 2009-03-31 17:14:32 Definition motion of the center of mass ellipse trajectory % this is the default PlanetPhysics preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here Let us consider a body with \PMlinkescapetext{mass} $m$ in a gravitational force \PMlinkname{field}{VectorField} exerted by the origin and directed always from the body towards the origin.\, Set the plane through the origin and the velocity vector $\vec{v}$ of the body.\, Clearly the body is forced to move constantly in this plane.\, Equip the plane with polar coordinate system $r,\,\varphi$ and denote the position vector of the body by $\vec{r}$.\, Then the velocity vector is \begin{align} \vec{v} \;=\; \frac{d\vec{r}}{dt} \;=\; \frac{d}{dt}(r\vec{r}^{\,0}) \;=\; \frac{dr}{dt}\vec{r}^{\,0}+r\frac{d\varphi}{dt}\vec{s}^{\,0}, \end{align} where $\vec{r}^{\,0}$ and $\vec{s}^{\,0}$ are the unit vectors in the direction of $\vec{r}$ and of $\vec{r}$ rotated 90 degrees anticlockwise ($\vec{r}^{\,0} = \vec{i}\cos\varphi+\vec{j}\sin\varphi$,\, whence\, $\frac{\vec{r}^{\,0}}{dt} = (-\vec{i}\sin\varphi+\vec{j}\cos\varphi)\frac{d\varphi}{dt} = \frac{d\varphi}{dt}\vec{s}^{\,0}$).\, Thus the kinetic energy of the body is $$E_k \;=\; \frac{1}{2}m\left(\frac{d\vec{r}}{dt}\right)^2 \;=\; \frac{1}{2}m\left(\left(\frac{dr}{dt}\right)^2+\left(r\frac{d\varphi}{dt}\right)^2\right)\!.$$ Because the gravitational force on the body is exerted along the position vector, its moment is 0 and therefore the angular momentum $$\vec{L} \;=\; \vec{r}\!\times\!m\frac{d\vec{r}}{dt} \;=\; mr^2\frac{d\varphi}{dt}\vec{r}^{\,0}\!\times\!\vec{s}^{\,0}$$ of the body is constant; thus its magnitude is a constant, $$mr^2\frac{d\varphi}{dt} \;=\; G,$$ whence \begin{align} \frac{d\varphi}{dt} \;=\; \frac{G}{mr^2}, \end{align} The central force\, $\vec{F} := \frac{k}{r^2}$\, (where $k$ is a constant) has the potential \, $U(r) = -\frac{k}{r}$.\, Thus the total energy\, $E = E_k+U(r)$ of the body, which is constant, may be written $$E \;=\; \frac{1}{2}m\left(\frac{dr}{dt}\right)^2+\frac{1}{2}mr^2\left(\frac{G}{mr^2}\right)^2-\frac{k}{r} \;=\; \frac{m}{2}\left(\frac{dr}{dt}\right)^2+\frac{G^2}{2mr^2}-\frac{k}{r}.$$ This equation may be revised to $$\left(\frac{dr}{dt}\right)^2+\frac{G^2}{m^2r^2}-\frac{2k}{mr}+\frac{k^2}{G^2} \;=\; \frac{2E}{m}+\frac{k^2}{G^2},$$ i.e. $$\left(\frac{dr}{dt}\right)^2+\left(\frac{k}{G}-\frac{G}{mr}\right)^2 \;=\; q^2$$ where $$q \;:=\; \sqrt{\frac{2}{m}\left(E+\frac{mk^2}{2G^2}\right)}$$ is a constant.\, We introduce still an auxiliary angle $\psi$ such that \begin{align} \frac{k}{G}-\frac{G}{mr} \;=\; q\cos\psi, \quad \frac{dr}{dt} \;=\; q\sin\psi. \end{align} Differentiation of the first of these equations implies $$\frac{G}{mr^2}\cdot\frac{dr}{dt} \;=\; -q\sin\psi\frac{d\psi}{dt} \;=\; -\frac{dr}{dt}\cdot\frac{d\psi}{dt},$$ whence, by (2), $$\frac{d\psi}{dt} \;=\; -\frac{G}{mr^2} \;=\; -\frac{d\varphi}{dt}.$$ This means that\, $\psi = C\!-\!\varphi$, where the constant $C$ is determined by the initial conditions.\, We can then solve $r$ from the first of the equations (3), obtaining \begin{align} r \;=\; \frac{G^2}{km\left(1-\frac{Gq}{k}\cos(C-\varphi)\right)} \;=\; \frac{p}{1-\varepsilon\cos(\varphi-C)}, \end{align} where $$p \;:=\; \frac{G^2}{km}, \quad \varepsilon \;:=\; \frac{Gq}{k}.$$ The result (4) shows that the trajectory of the body is a conic section.\\ [Not ready . . .]